Linear Transformations

Linear Mapping

Definition: Let $V$ and $W$ be linear spaces over the same field $F$. A mapping $\mathcal{T}: V \rightarrow W$ is called a linear mapping satisfying

here $V$ is called the domain of $T$ and $W$ is called the codomain of $\mathcal{T}$.

Example: Let $V = W$ polynomials of degree less than $n$ in $S$; $\mathcal{T} = \frac{d}{ds}$

Solution: Let $p,q \in V$ and $\alpha_1,\alpha_2 \in F$ then ,
p(s) = $\sum_{i=0}^{n-1} a_is^i$ and q(s) = $\sum_{i=0}^{n-1} b_is^i$
$\alpha_1p(s) + \alpha_2q(s) = \sum_{i=0}^{n-1} (\alpha_1a_i + \alpha_2b_i)s^i$
$\frac{d}{ds}(\alpha_1p(s) + \alpha_2q(s)) = \sum_{i=0}^{n-1} (\alpha_1a_i + \alpha_2b_i)is^{i-1} = \alpha_1\sum_{i=0}^{n-1} a_is^{i-1} + \alpha_2\sum_{i=0}^{n-1} b_is^{i-1} = \alpha_1\frac{dp}{ds} + \alpha_2\frac{dq}{ds}$
$\mathcal{T}(\alpha_1p+\alpha_2q) = \frac{d}{ds}(\alpha_1p+\alpha_2q) = \alpha_1\frac{dp}{ds} + \alpha_2\frac{dq}{ds} = \alpha_1\mathcal{T}(p) + \alpha_2\mathcal{T}(q) \ \blacksquare$

Example: Let $V = W = \mathbb{R^2}$. Let $\mathcal{A}$ be defined as,

Solution: Let $a,b \in F$ and $x_1,x_2 \in X$ with $x_1 = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}$ and $x_2 = \begin{bmatrix} \beta_1 \\ \beta_2 \end{bmatrix}$ then,
$\mathcal{A}(ax_1 + bx_2) = \mathcal{A}(a\begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} + b\begin{bmatrix} \beta_1 \\ \beta_2 \end{bmatrix}) = \mathcal{A}(\begin{bmatrix} a\alpha_1 \\ a\alpha_2 \end{bmatrix} + \begin{bmatrix} b\beta_1 \\ b\beta_2 \end{bmatrix}) = \mathcal{A}(\begin{bmatrix} a\alpha_1 + b\beta_1 \\ a\alpha_2 + b\beta_2 \end{bmatrix}) = \begin{bmatrix} a\alpha_1 + b\beta_1 \\ a\alpha_1 + a\alpha_2 + b\beta_1 + b\beta_2 \end{bmatrix} = a\begin{bmatrix} \alpha_1 \\ \alpha_1 + \alpha_2 \end{bmatrix} + b\begin{bmatrix} \beta_1 \\ \beta_1 + \beta_2 \end{bmatrix} = a\mathcal{A}(x_1) + b\mathcal{A}(x_2) \ \blacksquare$

Example: Let $V = W = \mathbb{R}$. is $\ \mathcal{A}x = (1-x)$ linear or not ?

Solution: Let $a,b \in F$ and $x_1,x_2 \in X$ then,

Rotation transformations in $\mathbb{R}^2$ are linear transformations.
Integration and differentiation are linear transformations.

Definition: Given a linear mapping $\mathcal{T}: V \rightarrow W$, the set of all vectors $x \in V$ such that $\mathcal{T}(x) = 0_W$ is called the null space of $\mathcal{T}$ and is denoted by $N(\mathcal{T})$. That is,

Definition: Given a linear mapping $\mathcal{T}: V \rightarrow W$, the set of all vectors $w \in W$ such that $w = \mathcal{T}(v)$ for some $v \in V$ is called the range of $\mathcal{T}$ and is denoted by $R(\mathcal{T})$. That is,

Claim: For a given linear mapping $\mathcal{T}: V \rightarrow W$, $N(\mathcal{T})$ is a linear subspace of $V$.

Proof: Let $x_1,x_2 \in N(\mathcal{T})$ and $a \in F$ show,

(S1). $x_1 + x_2 \in N(\mathcal{T})$
(S2). $ax_1 \in N(\mathcal{T})$

1- $\mathcal{T}(x_1 + x_2) = \mathcal{T}(x_1) + \mathcal{T}(x_2) = 0_W + 0_W = 0_W \implies x_1 + x_2 \in N(\mathcal{T})$
2- $\mathcal{T}(ax_1) = a\mathcal{T}(x_1) = a0_W = 0_W \implies ax_1 \in N(\mathcal{T}) \ \blacksquare$

Claim: For a given linear mapping $\mathcal{T}: V \rightarrow W$, $R(\mathcal{T})$ is a subspace of $W$.

Proof: Let $x_1,x_2 \in R(\mathcal{T})$ and $a \in F$ show,

(S1). $x_1 + x_2 \in R(\mathcal{T})$
(S2). $ax_1 \in R(\mathcal{T})$

Definition: A linear transformation $\mathcal{T}: V \rightarrow W$ is called one-to-one if $x_1 \neq x_2$ implies $\mathcal{T}(x_1) \neq \mathcal{T}(x_2)$ for all $x_1,x_2 \in V$.

Theorem: Let $\mathcal{T}: V \rightarrow W$ be a linear transformation. Then mapping $\mathcal{T}$ is one-to-one if and only if $N(\mathcal{T}) = \{0_V\}$.

Proof: We will prove the statement by contrapositive. Since it is an if and only if statement, we will prove both directions.

(Bacward direction) Assume that $N(\mathcal{T}) = \{0_V\}$ and $\mathcal{T}(x_1) = \mathcal{T}(x_2)$ for some $x_1,x_2 \in V$. Then,
$\mathcal{T}(x_1) - \mathcal{T}(x_2) = 0_W$
$\mathcal{T}(x_1 - x_2) = 0_W$
$x_1 - x_2 \in N(\mathcal{T})$
$x_1 - x_2 = 0_V$
$x_1 = x_2$
$\mathcal{T}$ is one-to-one.

(Forward direction) Assume that $\mathcal{T}$ is one-to-one and $x \in N(\mathcal{T})$. Then,
$\mathcal{T}(x) = 0_W$
$\mathcal{T}(0_V) = 0_W$
$x = 0_V$
$N(\mathcal{T}) = \{0_V\}$ $\blacksquare$

Definition: A linear transformation $\mathcal{T}: V \rightarrow W$ is called onto if $R(\mathcal{T}) = W$, otherwise if $R(\mathcal{T}) \subset W$ then $\mathcal{T}$ is called into.

Example: Let $V := \{f:[0, 1] \rightarrow \mathbb{R} \text{ and f is integrable} \}$. A transformation $\mathcal{A}: V \rightarrow \mathbb{R}$ is defined as,

Solution: Integration operation resulting in one-to-one transformation probably not true. Hence we can exploit the fact that the integration might result in zero.

Let $f(s) = 2s-1$ then,

$\mathcal{A}(f(s)) = \int_{0}^{1} (2s-1)ds = \left[s^2-s\right]_{0}^{1} = 0$
$\mathcal{A}(f(s)) = 0$
Then $\mathcal{A}(0) = 0_w$ and $\mathcal{A}(f(s)) = 0_w$ for some $f(s) \neq 0$.
$\mathcal{A}$ is not one-to-one.

moreover,

Let $f(s) = a then,

$\mathcal{A}(f(s)) = \int_{0}^{1} a ds = \left[as\right]_{0}^{1} = a$
Shows that $\mathcal{A}$ is onto.

Linear Transformations

Matrix Representations

Definition: Let $\mathcal{T}: V \rightarrow W$ be a linear transformation with $dim(V) = n$ and $dim(W) = m$. Let $\mathcal{B} = \{v_1,v_2,...,v_n\}$ be a basis for $V$ and $\mathcal{C} = \{w_1,w_2,...,w_m\}$ be a basis for $W$. Then, the matrix representation of $\mathcal{T}$ with respect to $\mathcal{B}$ and $\mathcal{C}$ is the $m \times n$ matrix $A$ such that,

Remark: The matrix representation of $\mathcal{T}$ with respect to $\mathcal{B}$ and $\mathcal{C}$ is denoted by $[\mathcal{T}]_{\mathcal{B}}^{\mathcal{C}}$.

Now we have a transformation represented as,

A formal procedure to obtain the matrix representation of a linear transformation

1. Take each basis vector $v_j$ in $\mathcal{B}$
2. Apply $\mathcal{A}$ to $v_j$ : $\mathcal{A}(v_j)$
3. Express the result in terms of the basis vectors in $\mathcal{C}$ : $\mathcal{A}(v_j) = \sum_{i=1}^{m} a_{ij}w_i$
4. The $j$th column of $[\mathcal{A}]_{\mathcal{B}}^{\mathcal{C}}$ is the vector $\begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix}$

Example: $V = \{Polynomials \ of \ degree \ less \ than \ 3\}$ and $W = \{Polynomials \ of \ degree \ less \ than \ 2\}$
Let $\mathcal{A}: V \rightarrow W$ be defined as,

Find the matrix representation of $\mathcal{A}$ with respect to the bases $\mathcal{B} = \{1, 1+s, 1+s+s^2, 1+s+s^2+s^3\}$ and $\mathcal{C} = \{1, 1+s, 1+s+s^2\}$.

Solution: $[w]_c = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} \ \text{and} \ [v]_b = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$

The full matrix representation of $\mathcal{A}$ is,

Example: Let $V = \mathbb{R}^2$ and $\mathcal{A}: V \rightarrow V$ be defined as,

Find the matrix representation of $\mathcal{A}$ with respect to the bases

$\mathcal{B} = \{\begin {bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin {bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin {bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin {bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\}$. and

$\mathcal{C} = \{\begin {bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin {bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin {bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}, \begin {bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\}$.

Solution: $[w]_c = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{bmatrix} \ \text{and} \ [v]_b = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$

Linear Transformations

Change of Basis

Definition: Let $\mathcal{B} = \{v_1,v_2,...,v_n\}$ and $\mathcal{C} = \{w_1,w_2,...,w_n\}$ be two bases for a linear space $V$. The change of basis matrix from $\mathcal{B}$ to $\mathcal{C}$ is the $n \times n$ matrix $P$ such that,

We know that a change of basis is a linear transformation. Hence,

in codomain perspective,

Example:

Solution: First we will find the change of basis in the domain matrix from $\mathcal{B}$ to $\mathcal{\bar{B}}$. That is more clearly stated as,

and given $[v]_B = P [v]_{\bar{B}}$, $\bar{A}$ is equal to,

In order to find $P$ we need to write the basis vectors in $\mathcal{\bar{B}}$ in terms of $\mathcal{B}$.

now $\bar{A}$ is equal to,

Now we will change the basis in the codomain to canonical basis while keeping the basis of the domain as also in canonical form. That is,

In order to find $Q^{-1}$ in a single step, we can write the basis vectors in $\mathcal{C}$ in terms of $\mathcal{\bar{C}}$.

as the final steps,

Given the matrix representation of a linear transformation $\mathcal{A}:V \rightarrow W$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$, one can draw the following diagram.

flowchart LR


B["[v]_b"] & A["[v]_b*"] -.- C{"v"}

B --[A]--> F

C --A--> D{"w"}
D --A'--> C

D -.- E["[w]_c*"] & F["[w]_c"]

A --[Q`AP]--> E 


A <--[P]--> B

E <--[Q]--> F

#EE501 - Linear Systems Theory at METU